What is expand (3r+10)^+(3r+3)^=(5r-2) ?

The solution to expand (3r+10)^+(3r+3)^=(5r-2) is 5r(3r+10)^+(3r+3)^=-2(3r+10)^+(3r+3)^=

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expand (3r+10)^+(3r+3)^=(5r-2)

expand

(3 r + 10)^{+} (3 r + 3)^{=} (5 r - 2)

5 r (3 r + 10)^{+} (3 r + 3)^{=} - 2 (3 r + 10)^{+} (3 r + 3)^{=}

Solution steps

(3 r + 10)^{+} (3 r + 3)^{=} (5 r - 2)

= (5 r - 2) (3 r + 10)^{+} (3 r + 3)^{=}

Apply the distributive law:

a (b - c) = ab - a c

a = (3 r + 10)^{+} (3 r + 3)^{=}, b = 5 r, c = 2

= (3 r + 10)^{+} (3 r + 3)^{=} \cdot 5 r - (3 r + 10)^{+} (3 r + 3)^{=} \cdot 2

= 5 r (3 r + 10)^{+} (3 r + 3)^{=} - 2 (3 r + 10)^{+} (3 r + 3)^{=}

\frac{0.5 z}{( z - 0.5 ) ( z - 0.7 ) ^{2}}

\frac{s - 2}{( s - 3 ) ( s ^{2} - 1 ) ( s ^{2} + 4 )}

\frac{3 s}{( s + 1 ) ^{4}}

\frac{e ^{- 4 x} ( e ^{8 x} - 1 )}{2}

\frac{1. 1 ^{2}}{- 1.1 + 1}

What is expand (3r+10)^+(3r+3)^=(5r-2) ?
The solution to expand (3r+10)^+(3r+3)^=(5r-2) is 5r(3r+10)^+(3r+3)^=-2(3r+10)^+(3r+3)^=